Exercise 5 Let \( f(x)=1+\frac{1}{x} \) and \( g(x)=\frac{1}{x} \). Find the following: a) \( (f+g)(x) \) and its domain. b) \( (f-g)(x) \) and its domain. c) \( (f g)(x) \) and its domain. d) \( \left(\frac{f}{g}\right)(x) \) and its domain. e) \( (f+g)(3) \) f) \( (f g)(2) \) g) \( (f \circ g)(x) \) and its domain. h) \( (g \circ f)(x) \) and its domain.

To solve these exercises, let's break down each part step by step. First, recall that \( f(x) = 1 + \frac{1}{x} \) and \( g(x) = \frac{1}{x} \). ### a) \( (f+g)(x) \) \[ (f+g)(x) = f(x) + g(x) = \left(1 + \frac{1}{x}\right) + \frac{1}{x} = 1 + \frac{2}{x} \] **Domain:** The domain of \( (f+g)(x) \) is all real numbers except \( x=0 \) (since both functions include \( \frac{1}{x} \)). So, Domain: \( x \in \mathbb{R} \setminus \{0\} \). ### b) \( (f-g)(x) \) \[ (f-g)(x) = f(x) - g(x) = \left(1 + \frac{1}{x}\right) - \frac{1}{x} = 1 \] **Domain:** The domain is also all real numbers except \( x=0 \) (since \( g(x) \) includes \( \frac{1}{x} \)). So, Domain: \( x \in \mathbb{R} \setminus \{0\} \). ### c) \( (fg)(x) \) \[ (fg)(x) = f(x) \cdot g(x) = \left(1 + \frac{1}{x}\right) \cdot \left(\frac{1}{x}\right) = \frac{1}{x} + \frac{1}{x^2} \] **Domain:** The domain remains the same as \( x \neq 0 \) since both \( f(x) \) and \( g(x) \) are undefined there. So, Domain: \( x \in \mathbb{R} \setminus \{0\} \). ### d) \( \left(\frac{f}{g}\right)(x) \) \[ \left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} = \frac{1 + \frac{1}{x}}{\frac{1}{x}} = \frac{(1 + \frac{1}{x}) \cdot x}{1} = x + 1 \] **Domain:** The domain is restricted by \( g(x) \neq 0 \), thus \( x \neq 0 \). So, Domain: \( x \in \mathbb{R} \setminus \{0\} \). ### e) \( (f+g)(3) \) \[ (f+g)(3) = 1 + \frac{2}{3} = 1 + 0.666... = \frac{5}{3} \] ### f) \( (fg)(2) \) \[ (fg)(2) = \frac{1}{2} + \frac{1}{2^2} = \frac{1}{2} + \frac{1}{4} = \frac{2}{4} + \frac{1}{4} = \frac{3}{4} \] ### g) \( (f \circ g)(x) \) \[ (f \circ g)(x) = f(g(x)) = f\left(\frac{1}{x}\right) = 1 + \frac{1}{\frac{1}{x}} = 1 + x \] **Domain:** The domain is restricted by \( g(x) \neq 0 \), hence \( x \neq 0 \). So, Domain: \( x \in \mathbb{R} \setminus \{0\} \). ### h) \( (g \circ f)(x) \) \[ (g \circ f)(x) = g(f(x)) = g\left(1 + \frac{1}{x}\right) = \frac{1}{1 + \frac{1}{x}} = \frac{x}{x+1} \] **Domain:** The domain is determined by \( f(x) \) being defined and must not be equal to -1. Thus \( x \neq 0 \). So, Domain: \( x \in \mathbb{R} \setminus \{0\} \). In conclusion, we have calculated the various operations and their domains accordingly!